Classic Genetics
AP Biology
Genetics is about using the seen, the phenotype, to infer the unseen, the genotype. Scientists make
predictions about modes of inheritance based on these inferences, and then compare the actual results
with these expectations. Although at a glance experimental results may seem to support the prediction,
best practice is to check the validity of data using a statistical test called Chi-square. Chi-square
evaluates a hypothesis by determining whether or not the differences between observed and predicted
results are due to chance or if they in fact represent a real difference. In statistics a difference is
‘significant’ if there is less than a 5% probability that the variation is due to random chance; if so, the
hypothesis is rejected.
Gregor Mendel initially considered a single trait of the pea at a time and from that deduced the Law
of Dominance. Later, he investigated whether the inheritance of one trait influenced another, what he
called the law of Independent Assortment. This involved predicting probabilities for multiple traits –
for example, pea plants that are tall and produce yellow seeds. Multi-trait crosses have exponentially
more outcomes, making a Punnett unwieldy; a better strategy is to determine the probability of each
trait, and then multiply those probabilities. The Fast Plants seeds you grew last semester were
dihybrid for leaf color and stem pigmentation.
Later, Thomas Morgan replicated Mendel’s experiments using fruit flies. Morgan found that the flies’
traits were not inherited independent of one another when the genes were linked, or found on the same
chromosome. The relative distance between two genes can be calculated from the offspring
frequencies and a gene ‘map’ generated. The farther apart the genes are, the greater the likelihood
crossing over will separate them.
A. MENDEL’S PEA DATA
Mendel first ran test crosses – pure dominant parent plants crossed with pure recessive.
Then he crossed the results of the first generation (f1) with the same.
1. Calculate the total number of plants Mendel observed for any two traits. Using that,
determine the predicted (expected) results for each trait. Recall that for traits such as
these – only 2 alleles, one dominant, one recessive – the dominant phenotype is
expected 75% of the time, the recessive 25%. Verify results with a partner. Record in
Table 1.
Parental traits
Dominant
Round seeds
Purple flowers
Inflated pods
Green pods
Axial flowers
Tall stem
Recessive
Wrinkled
White
Constricted
Yellow
Terminal
Dwarf
Actual
Dominant
Actual
Recessive
5474
705
882
428
651
787
1850
224
299
152
207
277
2. Since data almost always varies from the expected, we need to do a statistical test to see
if the difference is significant – if it is, we would have to reject the hypothesis. We do
this using the chi-square analysis, which is essentially an error analysis. Using the
formula below, compare your actual results with the predicted. Record all calculations.
Complete Table 1.
(Actual dominant – expected dominant)2
Chi = ----------------------------------------------expected dominant
2
3.
+
(actual recessive – expected rec. )2
------------------------------------expected recessive
Look up your chi-square result in Table 2 as follows:
• A chi-square result equal to no more than the 5% value means the observed
differences between data and hypothesis are expected to occur due to chance alone at
least 5% of the time. The closer the actual and expected are to each other, the greater
likelihood the differences would be expected. If so, the hypothesis is accepted.
In contrast, a chi-square larger than the value in the 5% column means the observed
results are so different from the predicted we must conclude the prediction is wrong.
• The degrees of freedom is one less than the number of groups. In this exercise there
are two phenotypes (groups), therefore 1 degree of freedom (df). Find the values your
chi square result falls between for 1 degree of freedom.
Are your predictions supported by the data? Determine the nearest percent probability
the difference between your hypothesis and results is due to chance. The higher this %
the more confident we are in our hypothesis, but scientists accept any chi-square value
that meets the 5% threshold.
TABLE 2 – CHI2 RESULTS
95%
90%
80%
70%
50%
30%
20%
10%
5%
df
1
0.004
0.02
0.064
0.15
0.46
1.07
1.64
2.71
3.84
2
0.10
0.21
0.45
0.71
1.39
2.41
3.22
4.60
5.99
3
0.35
0.58
1.00
1.42
2.37
3.66
4.64
6.25
7.82
B. FAST PLANT DIHYBRID DATA
Carolina Biological, who developed the seeds, hypothesizes that green leaves are
dominant to yellow, and anthocyanin pigment (creating purple stems) is dominant to no
pigment (green stems). The f1 seed I purchase is heterozygous (GgAa) for both genes; we
can test Carolina’s hypothesis against F2 data collected last fall.
4. From parents that are heterozygous for both green leaves and anthocyanin pigment,
determine all the possible genotypes of offspring. (Brainstorm or use foil or two
punnetts) Sum those that produce the same phenotype (green with anthocyanin; green,
no anthocyanin; yellow, anthocyanin; yellow, no anthocyanin) and determine the
predicted frequency of each. Multiply each predicted frequency by the number of plants
the class grew and surveyed. Record in Table 3.
5. Actual data of each plant type from years of class results is found in Table 4. Use the
chi-square formula to compare actual to expected for the 2012 data, for all years
combined, and for each of the traits separately. Consult Table 2 to check the validity of
the hypothesis.
C. MORGANS FLIES : LINKED GENES
Linkage is determined by crossing a hybrid organism with one that is recessive for both traits
in a backcross. Recombination information regarding three genes on the same arm of the
same chromosome is summarized in Table 4. Determine the relative location of each of gene.
6. Determine the expected frequencies of each phenotype from a cross of a heterozygote
and a homozygous recessive if
• the genes sort independently
• the genes are on the same chromosome (linked)
7. Do Morgan’s data match either expected frequencies? Recombination of the parent
pairs occurred but infrequently, indicating linked genes that occasionally are broken
up by crossing over. The closer the genes are to one another on the chromosome, the
more the traits show up together. The further apart, the more likely genetic
recombination occurs.
8. Determine the percent recombination – phenotypes that do not match either parent –
between each pair of traits. To find this value, use the following formula:
recombined phenotypes
---------------------------- x 100 = percent recombination
total offspring
9.
Every 1% of recombination equals 1 Linkage Map Unit (LMU) between two traits.
Use the three recombination percents to reflect the distances between the three genes
on a chromosome. Sequence them on a number line to reflect their positions relative to
one another.
TABLE 1 : MENDELS’ MONOHYBRID CROSS RESULTS
Parental traits
Dominant
Actual
Dominant
Actual
Recessive
Total
plants
Expected
Dominant
(.75)
Expected
recessive
(.25)
Chisquare
result
% likely the
differences are
due to chance
Recessive
Wrinkled
White
Constricted
TABLE 4: MORGAN’S DATA dominant traits are in bold type.
Black body & normal wing
X
Gray body & vestigial wing
Offspring
genotype
Expected
frequency
-no
linkage
Actual
result
Expected
result
Purple eye & normal wing
x
Red eye & vestigial wing
Black body & purple eye
x
Gray body & red eye
BbVv Bbvv bbVv bbvv RrVv Rrvv rrVv rrvv
BbRr Bbrr bbRr bbrr
0.25
0.25
0.25
.025
0.25
0.25
0.25
0.25
0.25
965
206
185
944
1339
151
154
1195 1159
0.25
0.25
0.25
67
78
1017
TABLE 1 : MENDELS’ MONOHYBRID CROSS RESULTS
Parental traits
Dominant
Actual
Dominant
Actual
Recessive
Total
plants
Expected
Dominant
(.75)
Expected
recessive
(.25)
Chisquare
result
% likely the
differences are
due to chance
Recessive
Wrinkled
White
Constricted
TABLE 4: MORGAN’S DATA dominant traits are in bold type.
Black body & normal wing
X
Gray body & vestigial wing
Offspring
genotype
Expected
frequency
-no
linkage
Actual
result
Expected
result
Purple eye & normal wing
x
Red eye & vestigial wing
Black body & purple eye
x
Gray body & red eye
BbVv Bbvv bbVv bbvv RrVv Rrvv rrVv rrvv
BbRr Bbrr bbRr bbrr
0.25
0.25
0.25
.025
0.25
0.25
0.25
0.25
0.25
965
206
185
944
1339
151
154
1195 1159
0.25
0.25
0.25
67
78
1017
KEY: MORGANS FLIES : LINKED GENES
Linkage is determined by crossing a hybrid organism with one that is recessive for both traits
in a backcross. Recombination information regarding three genes on the same arm of the
same chromosome is summarized in Table 2. Determine the relative location of each of gene.
1. Determine the expected frequencies of each phenotype from a cross of a heterozygote
and a homozygous recessive if
a) the genes sort independently (.25 each)
b) the genes are adjacent on a chromosome (linked) (0.5 each)
Table 2
dominant alleles are in bold
type.
Black body & normal wing
X
Gray body & vestigial wing
Purple eye & normal wing
x
Red eye & vestigial wing
Black body & purple eye
x
Gray body & red eye
BbVv Bbvv
bbVv
bbvv
RrVv
Rrvv
rrVv
rrvv
BbRr
Bbrr
bbRr
bbrr
965
206
185
944
1339
151
154
1195
1159
67
78
1017
.09
.08
.41
etc
.42
2. Do Morgan’s data match either expected frequencies? No. Recombination of the
parent pairs occurred but infrequently, indicating linked genes that occasionally are
broken up by crossing over. The closer the genes are to one another on the
chromosome, the more the traits show up together. The further apart, the more likely
genetic recombination occurs.
3. Determine the percent recombination – phenotypes that do not match either parent –
between each pair of traits. To find this value, use the following formula:
recombined phenotypes
---------------------------- x 100 = percent recombination
total offspring
ex: [(206 + 185)/2300] 100 = 16.9% recombination
DISCUSSION:
1. In the garden pea, yellow cotyledon color is dominant to green, and inflated pod shape is
dominant to the constricted form. Considering both of these traits jointly in self-fertilized
dihybrids, the progeny appeared in the following numbers:
193 green, inflated
184 yellow constricted
556 yellow, inflated
61 green, constricted
• Do these genes assort independently? Support your answer using Chi-square analysis.
DISCUSSION:
1. Given parents AABbCc and AabbCc, how likely is the offspring Aabbcc? AABbCc?
2. Two grey cats have a litter of 8 kittens. 3 kittens are grey, 3 are white and 2 are black.
• What genotypes might account for all these different colored kitties? Predict the
genotype of the parent cats, and the probability of black, grey and white kittens in a litter of
8.
• Does this example follow Mendel’s rules of dominance and recessiveness? Explain.
• Use the chi-square formula to test your prediction with the data from the litter.